Question

What is the frequency of the fundamental wave of a guitar string with a length of 1.5 m and a total mass of 60 g, tightened to a tension of 900 N?

Answer 1

To calculate the frequency of the fundamental wave of a guitar string, we can use the formula for the fundamental frequency (\( f_1 \)) of a vibrating string:

\[
f_1 = \frac{1}{2L} \sqrt{\frac{T}{\mu}}
\]

Where:
- \( L \) is the length of the string (in meters),
- \( T \) is the tension in the string (in newtons),
- \( \mu \) is the linear mass density of the string (in kg/m).

First, we need to calculate the linear mass density (\( \mu \)):

\[
\mu = \frac{m}{L}
\]

Where:
- \( m \) is the mass of the string (in kg),
- \( L \) is the length of the string (in meters).

Given:
- Length \( L = 1.5 \, \text{m} \)
- Mass \( m = 60 \, \text{g} = 0.06 \, \text{kg} \)
- Tension \( T = 900 \, \text{N} \)

Calculating the linear mass density:

\[
\mu = \frac{0.06 \, \text{kg}}{1.5 \, \text{m}} = 0.04 \, \text{kg/m}
\]

Now we can substitute \( L \), \( T \), and \( \mu \) into the fundamental frequency formula:

\[
f_1 = \frac{1}{2 \times 1.5} \sqrt{\frac{900}{0.04}}
\]

Calculating \( \frac{900}{0.04} \):

\[
\frac{900}{0.04} = 22500
\]

Now, we find the square root:

\[
\sqrt{22500} = 150
\]

Now substituting back to find \( f_1 \):

\[
f_1 = \frac{1}{3} \times 150 = 50 \, \text{Hz}
\]

Thus, the frequency of the fundamental wave of the guitar string is

\[
\boxed{50 \, \text{Hz}}
\]